Question

Verify the identity: (sin4x-sin2x)/(cos4x-cos2x) = TanX

Answer 1

is the question complete

Answer 2

ha oh yea. = tanx
sorry about that

Answer 3

i would try to be using the double angle identity

Answer 4

until no angles were doubled

Answer 5

there might be an easier identity but i cannot remember it
i would have to search for it

Answer 6

so far thats what i've been doing. I just struggle when I pick the wrong one.

Answer 7

\[\sin(4x)=\sin(2 \cdot 2x)=2 \cdot \sin(2x) \cos(2x)=2 \cdot 2\sin(x)\cos(x) [\cos^2(x)-\sin^2(x)]\]

Answer 8

maybe that will help

Answer 9

http://www.purplemath.com/modules/idents.htm
also the sum identities might work to

Answer 10

yea, its a good mixture of choosing between the sum/difference, the double angle or the pythgorean identities, i just suck at choosing which to use and at what case.

Answer 11

well if you never seen the sum identity talked about on their page
then i would go with double angle identity

Answer 12

since you have a ton of angles being doubled
and you are suppose to show it is tan(x) which requires no doubling

Answer 13

haha yea, i just realized the sum to product identities are in this chapter, they just aren't used in the example they refer to.

Answer 14

found a good explanation here.

Answer 15

yeah that was the sum identity they used

Answer 16

this is another approach here using double angles (it will be a little longer)
\[\frac{\sin(4x)-\sin(2x)}{\cos(4x)-\cos(2x)} \\ =\frac{2\sin(2x)\cos(2x)-\sin(2x)}{\cos^2(2x)-\sin^2(2x)-\cos(2x)}\]

I used the double angle on top and on bottom
Then you could be the bottom in terms of the trig function cos
then you can factor top and bottom
then something will cancel

Answer 17

\[=\frac{2\sin(2x)\cos(2x)-\sin(2x)}{\cos^2(2x)-(1-\cos^2(2x))-\cos(2x)}\\ \frac{2 \sin(2x)\cos(2x)-\sin(2x)}{\cos^2(2x)-1+\cos^2(2x)-\cos(2x)} \]
I used the Pythagorean identity there on bottom the bottom in terms of cosine
\[\frac{2\sin(2x)\cos(2x)-\sin(2x)}{2\cos^2(2x)-\cos(2x)-1}\]
factor top and bottom
and you will see something cancel afterwards

Answer 18

\[\frac{\sin(2x)(2\cos(2x)-1)}{2\cos^2(2x)-2\cos(2x)+\cos(2x)-1}\\=\frac{\sin(2x)(2\cos(2x)-1)}{2\cos(2x)[\cos(2x)-1]+1[\cos(2x)-1]} \\=\frac{\sin(2x)(2\cos(2x)-1)}{(\cos(2x)-1)(2\cos(2x)+1)}\]
well i lied I didn't see that 2 from before so nothing cancels yet

Answer 19

but you can write cos(2x)-1 as -2sin^2(x)
and write sin(2x)=2sin(x)cos(x) to see something cancel

Answer 20

but yeah the sum identities definitely look like a faster approach